Difference between revisions of "2012 IMO Problems/Problem 4"
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− | + | Find all functions <math>f: \mathbb{Z} \to \mathbb{Z}</math> such that, for all integers <math>a, b,</math> and <math>c</math> that satisfy <math>a + b+ c = 0</math>, the following equality holds: | |
+ | <cmath>f(a)^2 + f(b)^2 + f(c)^2 = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a).</cmath> | ||
+ | (Here <math>\mathbb{Z}</math> denotes the set of integers.) | ||
− | + | '''Solution:''' | |
− | a | + | Consider <math>a = b = c = 0.</math> Then <math>f(0)^2 + f(0)^2 + f(0)^2 = 2f(0)f(0) + 2f(0)f(0) + 2f(0)f(0) \Rightarrow 3f(0)^2 = 6f(0)^2 \Rightarrow</math> <cmath>f(0) = 0.</cmath> |
− | + | Now we look at <math>b = -a, c = 0.</math> <math>f(a)^2 + f(-a)^2 + f(0)^2 = 2f(a)f(-a) + 2f(-a)f(0) + 2f(0)f(a) \Rightarrow</math> <math>f(a)^2 + f(-a)^2 = 2f(a)f(-a) \Rightarrow</math> <math>f(a)^2 - 2f(a)f(-a) + f(-a)^2 = 0 \Rightarrow</math> <math>(f(a) - f(-a))^2=0 \Rightarrow</math> <cmath>f(a)=f(-a).</cmath> | |
− | + | What about <math>b = a, c = -2a?</math> Then <math>f(a)^2 + f(a)^2 + f(-2a)^2 = 2f(a)f(a) + 2f(a)f(-2a) + 2f(-2a)f(a) \Rightarrow</math> <math>2f(a)^2 + f(-2a)^2 = 2f(a)^2 + 4f(a)f(-2a) \Rightarrow</math> <math>f(-2a)^2 = 4f(a)f(-2a) \Rightarrow</math> <cmath>f(-2a) = f(2a) = 4f(a).</cmath> | |
− | + | We conjecture that <math>f(na) = n^2f(a).</math> Consider <math>b = n\cdot{a}, c = -(n+1)\cdot{a}</math> and assume that <math>f(a) \neq 0.</math> If it does, we get that the constant 0 function satisfies the conditions of the problem. | |
− | + | <math>f(a)^2 + f(n\cdot{a})^2 + f((n+1)\cdot{a})^2 =</math> <math>2f(a)f(n\cdot{a}) + 2f(n\cdot{a})f((n+1)\cdot{a}) + 2f((n+1)\cdot{a})f(a) \Rightarrow</math> | |
− | + | <math>f(a)^2 + (n^2f(a))^2 + ((n+1)^2f(a))^2 =</math> <math>2n^2f(a)^2 + 2n^2(n+1)^2f(a)^2 + 2(n+1)^2f(a)^2 \Rightarrow</math> | |
− | + | <math>1 + n^4 + (n+1)^4 = 2n^2 + 2n^2(n+1)^2 + 2(n+1)^2 \Rightarrow</math> | |
− | + | <math>1 + n^4 + n^4 + 4n^3 + 6n^2 + 4n + 1 = 2n^2 + 2n^4 + 4n^3 + 2n^2 + 2n^2 + 4n +2 \Rightarrow</math> | |
− | + | <math>2n^4 + 4n^3 + 6n^2 + 4n +2 = 2n^4 + 4n^3 + 6n^2 + 4n + 2 \hspace{7 mm} \checkmark</math> | |
− | + | <cmath>f(n\cdot{a}) = n^2f(a)</cmath> | |
− | + | We note that <math>f(a)=a^2f(1).</math> This means that we want to find what the possible values of <math>f(1)</math> are in order to finish the problem. <math>a + b + c = 0 \Rightarrow</math> <math>c = -(a+b) \Rightarrow</math> <math>a^4f(1)^2 + b^4f(1)^2 + (a+b)^4f(1)^2 =</math> <math>2a^2b^2f(1)^2 + 2b^2(a+b)^2f(1)^2 + 2(a+b)^2a^2f(1)^2.</math> Returning to the above manipulations (the ones used to show that <math>f(n\cdot{a}) = n^2f(a)</math>), we see that letting <math>n = \frac{a}{b}</math> and multiplying through by <math>b^4</math> yields precisely this result (again, assuming that <math>f(1) \neq 0,</math> with equality yielding the fact that the constant 0 function satisfies the condition). Therefore, <math>f(1)</math> can be any integer value (since <math>f</math> maps the integers to the integers only), and setting <math>f(1) = m</math>, we see that any function of the form <cmath>f(a) = m\cdot{a^2}, m \in \mathbb{Z}</cmath> satisfies the condition. | |
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Revision as of 20:02, 14 June 2013
Find all functions such that, for all integers and that satisfy , the following equality holds: (Here denotes the set of integers.)
Solution: Consider Then Now we look at What about Then We conjecture that Consider and assume that If it does, we get that the constant 0 function satisfies the conditions of the problem. We note that This means that we want to find what the possible values of are in order to finish the problem. Returning to the above manipulations (the ones used to show that ), we see that letting and multiplying through by yields precisely this result (again, assuming that with equality yielding the fact that the constant 0 function satisfies the condition). Therefore, can be any integer value (since maps the integers to the integers only), and setting , we see that any function of the form satisfies the condition.